Question: Let $R$ be the region in the fourth and first quadrants that is inside the polar curve $r=1+\cos^2(\theta)$ and outside the polar curve $r=\dfrac54$, as shown in the graph. The curves intersect at $\theta=-\dfrac{\pi}{3}$ and $\theta=\dfrac{\pi}{3}$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{-\scriptsize\dfrac{\pi}{3}}\left( 1+\cos^2(\theta)\right)^2\,d\theta+\dfrac{1}{2} \int_{\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{\pi}{2}}\dfrac{25}{16}\,d\theta$ (Choice B) B $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{\pi}{3}}\left( \left( 1+\cos^2(\theta)\right)^2-\dfrac{25}{16}\right)d\theta$ (Choice C) C $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{\pi}{2}}\left( \left( 1+\cos^2(\theta)\right)^2-\dfrac{25}{16}\right)d\theta$ (Choice D) D $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{3}}^{0}\left( 1+\cos^2(\theta)\right)^2\,d\theta+\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{3}}\dfrac{25}{16}\,d\theta$
Answer: Since we are dealing with two separate polar curves, a good first step is to identify two areas, each enclosed by a single curve, that together define $R$. Such are $R_1$ and $R_2$ : $y$ $x$ $ R_1$ $ R_2$ $ 1$ $ 1$ $R_1$ is enclosed by $r=1+\cos^2(\theta)$ and $R_2$ is enclosed by $r=\dfrac54$. Once we express them as integrals, we can find $R$ using the following relationship: $\text{Area of }R=\text{Area of }R_1-\text{Area of }R_2$ $R_1$ is enclosed by $r=1+\cos^2(\theta)$ between $\alpha=-\dfrac{\pi}{3}$ and $\beta=\dfrac{\pi}{3}$ : $\begin{aligned} \text{Area of }R_1&=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{\pi}{3}}\left( 1+\cos^2(\theta)\right)^2\,d\theta \end{aligned}$ $R_2$ is enclosed by $r=\dfrac54$ between $\alpha=-\dfrac{\pi}{3}$ and $\beta=\dfrac{\pi}{3}$ : $\begin{aligned} \text{Area of }R_2&=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{\pi}{3}}\left(\dfrac54\right)^2\,d\theta \\\\ &=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{\pi}{3}}\dfrac{25}{16}\,d\theta \end{aligned}$ Now we can express the area of $R$ : $\begin{aligned} &\phantom{=}\text{Area of }R \\\\ &=\text{Area of }R_1-\text{Area of }R_2 \\\\ &=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{\pi}{3}}\left( 1+\cos^2(\theta)\right)^2\,d\theta-\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{\pi}{3}}\dfrac{25}{16}\,d\theta \\\\ &=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{\pi}{3}}\left( \left( 1+\cos^2(\theta)\right)^2-\dfrac{25}{16}\right)d\theta \end{aligned}$